This is imo the issue with limits not being introduced properly at A-level. Limits don't simply involve substituting
h=0.
What
limh→0f(h)=y means is that as you pick values of
h closer and closer to 0 and substitute this into
f, you will get values closer to closer to
y.
In fact, you should be able to make
f(h) as close to
y as you want by picking a suitably close
h. For example, you can find a
h such that
f(h) is within
1001 of
y. Try this with, for example,
f(h)=h.
This edges towards a more formal definition of a limit which you don't quite need to understand at this level.
You can
only necessarily say:
h→0limg(h)f(h)=g(0)f(0)if a) both
f and
g are continuous (they are in this case, you don't have to worry about this) and b)
g(0)=0. But in this case,
h=0 on the denominator, so we can't just plug
0 in!
You want to start by simplifying the quotient:
hf(x+h)−f(x) (for example) as much as you can (within these simplifications we take
h=0), and then taking a limit. If you get something that's continuous at
h=0,
now you can just plug it in.
For example, if we have a function
f with
f(x)=x2 for all
x∈R we have for
h=0:
hf(x+h)−f(x)=h(x+h)2−x2=h2xh+h2=2x+h There is no issue here (no division by zeroes caused, etc.) with now plugging in
h=0 to get:
h→0lim(2x+h)=2x+0=2xgiving:
f′(x)=limh→0hf(x+h)−f(x)=2xNote that simplifying it to the point that we can just "plug
h=0" in isn't possible, sometimes it's more complicated. In this case we would aim to simplify the limit so we can apply other limits we already know.
Writing
lim throughout is poor notation but perfectly acceptable at this level.
This is a fairly complicated topic that confuses many undergrads so don't worry that you're struggling.